STRUCTURE OF ARGON ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. selleri , who gave me an award including a disc of the atomic philosopher Democritus. Argon (Ar) has 24 known isotopes, from Ar-30 to Ar-53 , three of which are stable, Ar-36, Ar-38, and Ar-40. On Earth, Ar-40 makes up 99.6% of natural argon. The longest-lived radioactive isotopes are Ar-39 with a half-life of 269 years, Ar-42 with a half-life of 32.9 years, and Ar-37 with a half-life of 35.04 days. All other isotopes have half-lives less than 2 hours, and most less than a minute. The least stable is Ar-30 with a half-life shorter than 20 nanoseconds. ' ' WHY Ar-36, Ar-38 AND Ar-40 WITH S = 0 ARE STABLE NUCLIDES ' After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. Since the stable Ar-36 with S = 0 consists of 18 protons and 18 neutrons '' ''in the following diagram of Ar-36 with S=0 we see that the n13, p13, n14, p14, n15 and p15 of the six horizontal planes like the +HP1 , -HP2 , +HP3, -HP4, +HP5 and -HP6 make vertical rectangles of S=0 with strong vertical bonds existing on the left side of the parallelepiped of Mg-24 with S=0 . (core). Also at the symmetrical positions the n16, p16, n17, p17, n18 and p18 make symmetrical vertical rectangles of S=0 with strong vertical bonds existing on the right side of the parallelepiped of Mg-24. ' DIAGRAM OF Ar-36 WITH S = 0 ' p12.....n12.....p18' ' -HP6 p15.......n11.......p11' ' . n10.....p10.....n18' ' +HP5 n15.......p9........n9 ' ' . p8.......n8........p17 ' ' -HP4 p14........n7........p7 ' ' . n6........p6.......n17' ' +HP3 n14 p5........n5 ' ' . p4........n4......p16' ' -HP2 p13........n3........p3 ' ' . n2........p2......n16' ' +HP1 n13.......p1........n1 ' However in the stable Ar-38 with S= 0 since the two extra neutrons like the p19 and p20 must make two bonds per neutron, we see that the stable structure of Ar-38 differs from that of the Ar-36. In the following diagram of Ar-38 we observe that the deuterons outside the parallelepiped of Mg-24 with S=0 like the p13n13, n14p14 and p17n17 give S =-1 and form a blank position for receiving the extra n19(+1/2) which exists at the fifth horizontal plane of positive spins (+HP5). Whereas the deuterons like the p15n15, n16p16 and p18n18 give S =+1 and form a blank position for receiving the n20(-1/2). Thus the two extra neutrons make two bonds per neutron and lead to the stability of Ar-38. Note that the total spin S=0 is given by S =0 -1 +1/2 +1 -1/2 = 0 '' '' For understanding the stable Ar-40 with 4 extra neutrons which make two bonds per neutron you can read my STRUCTURE OF Ar-40 AND K-39 . Note that the stability of Ar-40 is due to the fact that it has 4 extra neutrons with two bonds per neutron which contribute to the increase of the binding energies. Also the stability of Ar-38 is due to the fact that it has two extra neutrons with two bonds per neutron. ' DIAGRAM OF Ar-38 WITH S = 0' ' p12......n12' ' -HP6 n11.......p11' ' . n10.......p10.....n18' ' +HP5 n19.......p9.......n9.........p18 ' ' . n17........p8.........n8.......p16 ' ' -HP4 p17.......n7.......p7.......n16 ' ' . p14.......n6........p6.......n15' ' +HP3 n14.......p5........n5.......p15 ' ' . n13.......p4........n4' ' -HP2 p13.......n3.........p3.......n20 ' ' . n2........p2' ' +HP1 p1........n1 ' STRUCTURE OF Ar-34, Ar-32 AND Ar-30 WITH S =0 If we remove the neutrons n19 and n20 of the stable Ar-38 we will get a structure of 18 protons and 18 neutrons in which there are positions for removing more neutrons (at the corners of the parallelepiped and the rectangles formed by the deuterons). Whereas in the stable structure of Ar-36 there are not such positions for removing neutrons. Under this condition in the absence of neutrons with opposite spins the structure of the above nuclides is based on the structure of Ar-38. For example in the structure of Ar-30 with S=0 we see that there are 8 absent neutrons with opposite spins. NUCLEAR STRUCTURE OF Ar-42, Ar-44, Ar-46, Ar-48 Ar-50, AND Ar-52 WITH S=0 Using the diagram of Ar-40 we see that there are not more blank positions than those of the structure of Ar-40 which has 4 extra neutrons with two bonds per neutron. Thus, the structure of the above nuclides is based on the structure of Ar-40 in which we add extra neutrons of opposite spins making single bonds unable to overcome the nn repulsions. For example the Ar-52 with S=0 has 12 more extra neutrons with single bonds of opposite spins than those of Ar-40 with S=0. ' ' NUCLEAR STRUCTURE OF Ar-37, Ar-45, Ar-47, Ar-49 AND Ar-51 ''' In the diagram of Ar-37 with S =+3/2 we see that the deuterons n13p13, p14n14, n15p15, p16n16, n17p17 and p18n18 give a total S = +2,, while the blank position at the -HP2 receives the n19(-1/2). Thus the total spin is given by S = 0 +2 -1/2 = +3/2. Therefore the structure of Ar-45 with S = +1/2 is based on the structure of Ar-37. Adding two extra neutrons with single bonds of negative spins and 6 extra neutrons of opposite spins giving S=0 we get the structure of the unstable Ar-45 with S =+1/2 . That is S = 0 + 3/2 + 2(-1/2) + 0 = +1/2. In the case of Ar-47 with S = -3/2 we conclude that all nucleons of Ar-37 change the spins of the six horizontal planes. Thus we have -HP1, +HP2, -Hp3, +HP4, -HP5 and +HP6 giving S =-3/2. Under this new arrangement of spins the Ar-47 with S = -3/2 has 10 extra neutrons of opposite spins making single bonds. Also in the Ar-49 and Ar-51 with S=-3/2 we have 12 and 14 extra neutrons of opposite spins respectively. In the case of Ar-53 with S=-5/2 we have two extra neutrons of negative spins and 14 extra neutrons of opposite spins. '''STRUCTURE OF Ar-35, Ar-33, AND Ar-31 Using the structure of Ar-37 with S = +3/2 we see that in the absence of 2 neutrons of opposite spins we get the structure of Ar-35 with the same S =+3/2. However in the Ar-33 with S =+1/2 we have two absent neutrons of positive spins and 12 absent neutrons with opposite spins giving S=0. That is ,the total spin is given by S = +3/2 -2(+1/2) - 0 = +1/2 Finally using again the structure of Ar -37 we see that in the structure of Ar-31 with S = +5/2 we have two absent neutrons with negative spins and 14 absent neutrons of opposite spins giving S=0. That is, the total spin is given by S = +3/2 -2(-1/2) - 0 = +5/2 ' ' NUCLEAR STRUCTURE OF Ar-39, Ar-41, Ar-43, AND Ar-53 In the following diagram of Ar-39 we see that the core is not the parallelepiped of Mg-24. Here the parallelepiped has 5 horizontal planes like the -HP1, +HP2 -HP3 +HP4 and -HP5 giving S = -3. Also the deuterons p16n16n17p17 and p18n18 give S = -1 and make a blank position at +HP2 for receiving the n19(+1/2). Under this new arrangement of nucleons the two extra neutrons of single bonds like the n20(+1/2) and the n21(-1/2) contribute to the total spin as S = -3 -1 +1/2 +1/2 -1/2 = -7/2 Since the structure of Ar-41 with S=-7/2 is based on the structure of Ar-39 with S=-7/2 we conclude that it has two more extra neutrons of opposite spins. However the Ar-43 with S =-5/2 based on the same structure of Ar-37 has two extra neutrons of negative spins and 4 extra neutrons of opposite spins. Similarly the Ar-53 with S=-5/2 has 10 more extra neutrons of opposite spins than those of the structure of Ar-43. ' ' ' DIAGRAM OF Ar-39 WITH S = -7/2' ' p15.....n10......p10........n18' ' -HP5 n15.......p9.......n9........p18 ' ' . n14.......p8........n8.......p17...n20 ' ' +HP4 p14......n7.......p7......n17 ' ' . p13.......n6........p6.......n16' ' -HP3 n13.......p5........n5.......p16 ' ' . n12........p4........n4' ' +HP2 p12......n3........p3.......n19 ' ' . n21...p11......n2........p2' ' -HP1 n11.....p1.......n1 ' Category:Fundamental physics concepts